ACSI 2022 Y6 HL Mathematics Prelim Exam Paper 2 (Student Solutions)

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 1 PRELIM 2022 PAPER 2 YEAR 6 IB DIPLOMA PROGRAMME MATHEMATICS : ANALYSIS AND APPROACHES HIGHER LEVEL STUDENT SOLUTIONS SECTION A Qn Solution COMMENTS 1. [4 marks] (a) Area of the minor segment = Area of the sector – Area of OAB 2 21 1 sin2 2r r   Some mistook sector for segment. (b)   2 2 2 2 2 1 1 1 sin2 2 4 2 sin 2 2sin (shown) r r r r r                  This is correct if (a) is correct O A B

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 2 Qn Solution COMMENTS (c) 2.30988 2.31 (3sf)  Many could not do. Must use Graph or nsolve 2. [4 marks] ' sin cosf x x x Observe that  cos sind x xdx  .     1 2 3 2 3 2 sin cos ( sin ) cos cos 3 2 2 cos3 f x x x dx x x dx x c x c          Substituting , 0 :3x y  3 2 3 2 20 cos3 3 2 1 1 3 2 3 2 c c            Thus in exact form we have    3 22 1cos3 3 2f x x  Many did not see this as in the form of ∫𝑓ᇱ(𝑥)[𝑓(𝑥)]௡𝑑𝑥=[௙(௫)]೙శభ ௡ାଵ +𝑐 Not acceptable for decimal value of ଵ ଷ√ଶ. All other surd forms of ଵ ଷ√ଶ are accepted. 3. [6 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 3 Qn Solution COMMENTS (a)    0.95 0.6 0.7 0.35 P X Y P X P Y P X Y P X Y P X Y              Most got it correct (b)    0.35 0.6 7 12 P X YP Y X P X    ଻ ଵଶ is an exact value so no need to convert to 0.583. Some students only gave as 0.58. (c) 𝑃(𝑋∪𝑌ᇱ)=0.25+0.35+0.05=0.65 Alternatively, accept ' 1 1 0.7 0.35 0.65 P X Y P Y P X Y         Badly done. 4. [7 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 4 Qn Solution COMMENTS (a) From GDC we have Mean 50.3 Variance   22 23.7152 562.4107 562 (3sf)    Key in data and use one-var stats. Scroll down and the standard deviation is 23.7152 so need to square it to obtain the variance. Please correct to 3 significant figures (b) From GDC we have 33, 51, 73a b c   (c) New mean 50.3new q   New standard deviation 23.7 (3sf)new  Standard deviation measures the spread of the data so it is still the same if every datum is decreased by the same value. 5. [7 marks] (a) ln 1 v A t B dv dt t B       Substituting 110, : 20 dvt dt  1 1 20 10B   10B  Substituting 100, 0 :t v    0 ln 100 10 ln 110 A A      Most got it correct but some could not differentiate v.

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 5 Qn Solution COMMENTS (b) Distance = 2.10 m (3sf) 4th second means t = 3 to t = 4 because first second is t = 0 to t = 1 6. [5 marks] (a) Number of ways is 13! 6227020800 This is an exact answer. Do not correct to 6.23×10ଽ (b) Number of ways is 6 2 3! 8! 2903040    (6 possible configurations of 5 seats in a row, 2 position swaps for Mr & Mrs Lee, 3! - for the Lee children, 8! - for remaining members of tour group) Lees: 3!×2 Others in first row: 8C5 Permutation of first row: 6! Because LEES is one block. 2nd row: 3P3 Ans: 3!×2×8𝐶5×6!×3! 7. [9 marks] (a)   0 0 0 0 ln seclim lim sec tan seclim 1 lim tan 0 x x x x xf x x x x x x         Recall: ௗ ௗ௫ൣln൫𝑓(𝑥)൯൧ =௙ᇱ(௫) ௙(௫) Here f(x) is sec x whose differentiation is sec x tan x. Qn asks you to use l’Hopital’s rule so no need to show indeterminate. (b)          2 (3) 2 (4) 2 2 2 2 2 Let ln sec sec tan' tan sec '' sec 2 sec tan sec 2sec tan 4sec (sec tan ) tan 2sec sec 2sec 2 tan sec g x x x xg x x x g x x g x x x x x x g x x x x x x x x x x           Some used differentiation at a point on GDC to get all the specific derivatives at x = 0

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 6 Qn Solution COMMENTS (b) Substituting 0x into the above expressions:      2 (3) 2 (4) 2 2 2 0 ln sec0 ln1 0 ' 0 tan 0 0 '' 0 sec (0) 1 0 2sec (0) tan 0 0 0 2sec (0) 2 tan 0 sec 0 2 g g g g g                By Maclaurin’s series, the expansion is:    2 4 2 4 1 2ln sec ...2! 4! 1 1 ... (shown)2 12 g x x x x x x        (c) So the volume generated when the region is revolved around the 𝑥-axis is 2 20.9767 2 0 0.9767 3 ln(sec ) 2 0.525293 0.525 units (3sf) x dx x dxx                    Big conceptual error for some. How can you find area and multiply by 2π? Please do not use 𝑢ଷ for units3. 8. [6 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 7 Qn Solution COMMENTS (a) Euler’s method using GDC: When 1.3, 1.25 (3sf)x y  New y = Old y + ௗ௬ ௗ௫×𝑠𝑡𝑒𝑝 (b) 2d d yy x ex By variable separable we have 2 3 3 y y e dy x dx xe c         Sub in boundary condition 1, 0.5 :x y  0.5 0.5 1 3 1 0.939864 0.940 (3sf)3 e c c e           Thus the particular solution for the differential equation is 3 3 3 0.9403 ln 0.9403 ln 0.9403 y xe xy xy                Also accept 3 0.5 1ln 3 3 xy e       Find particular solution means find the equation for y with the value of c. Use the given condition x = 1, y = 0.5, not the approximated x = 1.3, y = 1.25 9. [7 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 8 Qn Solution COMMENTS (a) Well done except for not correcting to 3 significant figures Use linsolve in algebra menu.

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 9 Qn Solution COMMENTS (b) Let be the number of pineapples rejected out of pineapples in the crate. Then . We have . GDC type in Largest . Can use invbinomN. Check the matrix form. SECTION B Qn Solution COMMENTS 10. [15 marks] (a) Note final answers should be rounded to 3 significant figures (b) Points of intersection of the two graphs are Common error: Students were penalized for not sketching graph or showing any working or attempt at solving.

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 10 Qn Solution COMMENTS (c) is an increasing function when . Thus the range of values is Students should state explicitly that if they did not sketch the graph. Common error: Students were penalized if the inequality was not strict. (d) For points of inflexion, there is a change in concavity which can be graphically represented by a change of sign in the graph of : Thus the points of inflexion occur at Common error: Justification for point of inflexion was not explicitly stated even when the correct graph was plotted. In future, please state change in concavity / change in sign of 2nd derivative as seen from the graph or table.